 oliviayychengwh

2022-01-02

How to prove this inequality in interval: Thomas Nickerson

Expert

Restrict the domain to some meaningful interval such as $x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ so that $\mathrm{cos}x>0$. In that case, using the AM-GM inequality you have:
$LHS=\mathrm{cos}x+\mathrm{cos}x+{\mathrm{sec}}^{2}x-3\ge 3\sqrt{\mathrm{cos}x\cdot \mathrm{cos}x\cdot {\mathrm{sec}}^{2}x}-3=3-3=0=RHS$ Papilys3q

Expert

We have
$2\mathrm{cos}x+{\mathrm{sec}}^{2}x-3=2\left(\mathrm{cos}x-1\right)+\left(\frac{1}{{\mathrm{cos}}^{2}x}-1\right)=\frac{\left(2\mathrm{cos}x+1\right){\left(\mathrm{cos}x-1\right)}^{2}}{{\mathrm{cos}}^{2}x}$
Thefore the inequality does not hold. Vasquez

Expert

By Fermat's theorem, we examine zeroes of derivative of the function $f\left(x\right)=2\mathrm{cos}x+{\mathrm{sec}}^{2}x-3$ to find the extrema, that is, to solve the following:
$f\prime \left(x\right)=2\mathrm{tan}x{\mathrm{sec}}^{2}x-2\mathrm{sin}x=0$
The solution is $x=\left\{k\pi |k\in \mathbb{Z}\right\}$. When $x=\pi$, we get
$2\mathrm{cos}\left(x\right)+{\mathrm{sec}}^{2}\left(x\right)-3=-4$
Thus $2\mathrm{cos}x+{\mathrm{sec}}^{2}x-3\ge 0$ doesn't hold.
Since the interval is added, we can prove the claim is true. For x=0, f(x)=0. And when $x\in \left(0,\frac{\pi }{2}\right)$, it is trivial that $f\prime \left(x\right)\ge 0$, thus f(x) monotonously increases on $\left(0,\frac{\pi }{2}\right)$. Hence and we finished the proof.

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